# Step-by-step Solution

## Find the higher order derivative of $\sin\left(x^3+y^5\right)$

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asin
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### Videos

$6x\cos\left(x^3+y^5\right)-9x^{4}\sin\left(x^3+y^5\right)$

## Step-by-step explanation

Problem to solve:

$\frac{d^2}{dx^2}\left(\sin\left(x^3+y^5\right)\right)$
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Rewriting the high order derivative

$\frac{d}{dx}\left(\frac{d}{dx}\left(\sin\left(x^3+y^5\right)\right)\right)$
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The derivative of the sine of a function is equal to the cosine of that function times the derivative of that function, in other words, if ${f(x) = \sin(x)}$, then ${f'(x) = \cos(x)\cdot D_x(x)}$

$\frac{d}{dx}\left(\cos\left(x^3+y^5\right)\frac{d}{dx}\left(x^3+y^5\right)\right)$

$6x\cos\left(x^3+y^5\right)-9x^{4}\sin\left(x^3+y^5\right)$
$\frac{d^2}{dx^2}\left(\sin\left(x^3+y^5\right)\right)$